let-U-n-k-0-n-1-k-2-k-1-find-a-equivalent-of-U-n-n-

Question Number 74352 by mathmax by abdo last updated on 22/Nov/19

l e t U_{n} = \sum_{k = 0}^{n} \frac{1}{k^{2} + k + 1} f i n d a e q u i v a l e n t o f U_{n} (n \to + \infty)

Answered by mind is power last updated on 24/Nov/19

l e t f (z) = \frac{π}{z^{2} + z + 1} . c o t (z)

w e c a n f i n d t h e v a l u e o f t h e s o m

p o l s o f f a r e Z \cup {j, \bar{j}}

R e s (f, n \in Z) = \frac{1}{n^{2} + n + 1}

∣ f (z) ∣⩽ \frac{∣ c o t h (R e (z)) ∣}{∣ z^{2} + z + 1 ∣} \to 0 ∣ z ∣\to \infty

\Rightarrow \int_{C_{R}} f (z) d z = 0 R e s t h \Rightarrow 2 i π \sum_{n \in Z} R e s (f, n) + 2 i π R e s (f, j, \bar{j})

R e s (f, j) =

\frac{π}{(j - \bar{j})} . \frac{c o s (π j)}{s i n (π j)} = \frac{π}{i \sqrt{3}} . \frac{c o s (- \frac{π}{2} + i \frac{\sqrt{3}}{2})}{s i n (- \frac{π}{2} + i \sqrt{\frac{3}{4}})} = - \frac{π}{i \sqrt{3}} . \frac{s i n (0.5 i \sqrt{3})}{c o s (\frac{i \sqrt{3}}{2})}

- \frac{π}{\sqrt{3}} . t h (\frac{\sqrt{3}}{2})

R e s (f, \bar{j}) = \frac{π}{j - \bar{j}} . \frac{c o s (\frac{- π}{2} - \frac{i \sqrt{3}}{2})}{s i n (- \frac{π}{2} - i \frac{\sqrt{3}}{2})} = \frac{π}{i \sqrt{3}} . \frac{- s i n (\frac{i \sqrt{3}}{2})}{- c o s (\frac{i \sqrt{3}}{2})} = - \frac{π}{\sqrt{3}} t h (\frac{\sqrt{3}}{2})

\Rightarrow 2 i π \sum_{n \in Z} \frac{1}{1 + n + n^{2}} + 2 i π (- \frac{2 π}{\sqrt{3}} . t h (\frac{\sqrt{3}}{2})) = 0

\Rightarrow \sum_{n \in Z} \frac{1}{1 + n + n^{2}} = \frac{2 π}{\sqrt{3}} . t h (\frac{\sqrt{3}}{2})

\sum_{n \in N} \frac{1}{1 + n + n^{2}} + \sum_{n \in N^{*}} \frac{1}{1 - n + n^{2}} = \sum_{n \in Z} \frac{1}{1 + n + n^{2}}

\Rightarrow \sum_{n \in N^{*}} \frac{1}{1 - n + n^{2}} = \sum_{n \in N^{*}} \frac{1}{(n - 1) + 1 + {(n - 1)}^{2}} = \sum_{n \in N} \frac{1}{1 + n + n^{2}}

\Rightarrow 2 \sum_{n \in N} \frac{1}{1 + n + n^{2}} = \frac{2 π}{\sqrt{3}} . t h (\frac{\sqrt{3}}{2}) \Rightarrow \sum_{n \in N} \frac{1}{1 + n + n^{2}} = \frac{π}{\sqrt{3}} t h (\frac{\sqrt{3}}{2})

Commented by abdomathmax last updated on 24/Nov/19

t h a n k x s i r .

Commented by mind is power last updated on 24/Nov/19

y^{'} r e w e l c o m

Commented by mathmax by abdo last updated on 24/Nov/19

f (z) = \frac{π c o t a n (π z)}{z^{2} + z + 1}

Commented by mind is power last updated on 02/Dec/19

yes sir